The balanced equation for the preparation of NaBr is as follows:Fe+Br2→FeBr2 …(i) (yield: 100 %) 3FeBr2+Br2→Fe3Br8...(ii) (yield: 100 %)Fe3Br8+4Na2CO3→8NaBr+4CO2+Fe3O4…(iii) (yield: 60 %) Calculate the mass of iron required to produce 2.06×103 g of NaBr.(Considering Na2CO3 and Br2 to be present in excess)(molar mass of Fe=56 g/mol,Br=80 g/mol,Na=23 g/mol)

The balanced equation for the preparation of NaBr is as follows:Fe+Br2

1.	The balanced equation for the preparation of NaBr is as follows:Fe+Br2→FeBr2 …(i) (yield: 100 %) 3FeBr2+Br2→Fe3Br8...(ii) (yield: 100 %)Fe3Br8+4Na2CO3→8NaBr+4CO2+Fe3O4…(iii) (yield: 60 %) Calculate the mass of iron required to produce 2.06×103 g of NaBr.(Considering Na2CO3 and Br2 to be present in excess)(molar mass of Fe=56 g/mol,Br=80 g/mol,Na=23 g/mol)
Answer» The balanced equation for the preparation of $N a B r$ is as follows: $F e + B r_{2} \to F e B r_{2} \dots(i) (yield: 100 %)$ $3 F e B r_{2} + B r_{2} \to F e_{3} B r_{8} ...(ii) (yield: 100 %)$ $F e_{3} B r_{8} + 4 N a_{2} C O_{3} \to 8 N a B r + 4 C O_{2} + F e_{3} O_{4} \dots(iii) (yield: 60 %)$ Calculate the mass of iron required to produce $2.06 \times 10^{3} g$ of $N a B r$ . (Considering $N a_{2} C O_{3}$ and $B r_{2}$ to be present in excess) (molar mass of $F e = 56 g/mol, B r = 80 g/mol, N a = 23 g/mol$ )